Determine an expression for the area of $\triangle QCA$ in terms of $p$. Your answer should be simplified as much as possible. [asy]
size(5cm);defaultpen(fontsize(9));
pair o = (0, 0); pair q = (0, 12); pair b = (12, 0);
pair a = (2, 12); pair t = (2, 0); pair c = (0, 9);

draw((-2, 0)--(15, 0), Arrow);
draw((0, -2)--(0, 15), Arrow);
draw(q--a--b);
//draw(a--t);
draw(a--c--b);

label("$Q(0, 12)$", q, W);
label("$A(2, 12)$", a, NE);
label("$B(12, 0)$", b, S);
label("$O(0, 0)$", o, SW);
label("$x$", (15, 0), E);
label("$y$", (0, 15), N);
//label("$T(2, 0)$", t, S + 0.6 * E);
label("$C(0, p)$", c, W);
[/asy]
Since $QA$ is perpendicular to $QC$, we can treat $QC$ as the height of $\triangle QCA$ and $QA$ as the base.  The area of $\triangle QCA$ is $$\frac{1}{2}\times QA\times QC=\frac{1}{2}\times(2-0)\times(12-p)=\frac{1}{2}\times2\times (12-p)=\boxed{12-p}.$$